The total number of possible proper three digit integers that can be formed using 0,1,3,4,and 5 without repetation such that they are divisible by 5
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You should note that not every five-digit number made of 0,1,2,3,4,5 is divisible by 3.
A number is divisible by 3 if the sum of its digits is. Since 0+1+2+3+4+5=15, which is divisible by 3, and you need only 5 digits, you must leave our a number that's divisible by 3, namely 0 or 3.
If you leave out 0, you end up with five numbers, which can be ordered in 5!=120 ways.
If you leave out 3, you end up with 120 numbers again, but 15 of them would begin with 0 so they actually have 4 digits. This means you have to rule out 1205=24 numbers.
To sum up, you have 120+120−24=216 as you mentioned.
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